# Value of functions when approximating area using Riemann sums

When approximating the area beneath the curve $f(x)=x^2+1$ on the interval $[0,3]$ using a left Riemann sum with 4 rectangles, I calculated that the height of each rectangle, in order from left to right, would be

$f(0)$, $f(\frac{3}{4})$, $f(\frac{3}{2})$, $f(\frac{9}{4})$

and therefore, the area would be

$A=\frac{3}{4}\cdot f(0)+\frac{3}{4}\cdot f(\frac{3}{4})+\frac{3}{4}\cdot f(\frac{3}{2})+\frac{3}{4}\cdot f(\frac{9}{4})$ (I'll call this equation 1)

Now I know the next step is

$A=\frac{3}{4}\cdot1+\frac{3}{4}\cdot \frac{25}{16}+\frac{3}{4}\cdot \frac{13}{4}+\frac{3}{4}\cdot\frac{97}{16}=\frac{285}{32}\approx8.906$ (I'll call this equation 2)

But it's this step that confuses me.

How does one know for example that $f(\frac{3}{4})$ from equation 1 goes to $\frac{25}{16}$ in equation 2 and $f(\frac{9}{4})$ from equation 1 goes to $\frac{97}{16}$ in equation 2? (The same goes for the other variables).

I apologise if this is written in a confusing way.

-

Well $f(x) = x^2+1$

$$f\left(\frac 3 4\right) = \left(\frac 3 4\right)^2+1 = \frac 9 {16} + \frac {16}{16} = \frac {25} {16}$$ and $$f\left(\frac 9 4\right) = \left(\frac 9 4\right)^2+1 = \frac {81} {16} + \frac {16}{16} = \frac {97} {16}$$

-
Ah my goodness of course, how did I not see this, I feel so stupid! –  Olly Price Aug 15 '12 at 14:36
You are applying the function to the leftmost side of the rectangle to get the left sum. –  Mark Bennet Aug 15 '12 at 14:37

The values such as $\frac{25}{16}$ and $\frac{97}{16}$ were obtained by evaluating the original function $f(x) = x^2 +1$. Namely:

$f(0) = 0^2+1 = 1$

$f(\frac{3}{4}) = (\frac{3}{4})^2 + 1 = \frac{25}{16}$

$f(\frac{3}{2}) = (\frac{3}{2})^2 + 1 = \frac{13}{4}$

$f(\frac{9}{4}) = (\frac{9}{4})^2 + 1 = \frac{97}{16}$

Each value is used to estimate the height of the curve for a certain segment of the interval $[0,3]$.

For example, the value $f(0)=1$ is an estimate for the height of the curve for the first segment, $[0,\frac{3}{4}]$. So the area of the first rectangle is $\frac{3}{4} . 1 = \frac{3}{4}$

A similar calculation is done for all 4 rectangles, and then sum them up to get the approximate total area under the curve.

If you increase the number of rectangles in your approximation, you will generally get a better estimate for the area under the curve.

-